A rod of length 10 cm carries a charge of +50 C uniformly distributed along its length. Determine the magnitude of the electric field at a point located 10 cm from both ends of the rod.

Step-by-step solution

Length of the rod, L = 10 cm = 0.1 m Charge on the rod, Q = 50 C = 50 10⁻⁶ C Linear charge density, = Q L = 50 10⁻⁶ 0.1 = 5 10⁻⁴ C/m The point is at a distance of 10 cm from both ends of the rod. Thus, the point and the two ends of the rod form an equilateral triangle of side 10 cm . The point lies on the perpendicular bisector of the rod. The perpendicular distance of the point from the rod is r = 10^2 - 5^2 = 5 3 cm = 5 3 10⁻² m . The angle subtended by each half of the rod at the point is ₁ = ₂ = 30^ . The elect